ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 865

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\[1)\ y = x^{4}\text{\ \ }и\ \ y = x^{6} + 2x^{2}\]

\[x^{4} = x^{6} + 2x^{2}\]

\[x^{6} - x^{4} + 2x^{2} = 0\]

\[x^{2} \bullet \left( x^{4} - x^{2} + 2 \right) = 0\]

\[x = 0\]

\[x^{4} - x^{2} + 2 = 0\]

\[y = x^{2}:\]

\[y^{2} - y + 2 = 0\]

\[D = 1 - 8 = - 7 < 0.\]

\[1)\ y^{'}(x) = \left( x^{4} \right)^{'} = 4x^{3}\]

\[k = f^{'}(0) = 4 \bullet 0^{3} = 0\]

\[a = arctg\ 0 = 0.\]

\[2)\ y^{'}(x) = \left( x^{6} \right)^{'} + 2 \bullet \left( x^{2} \right)^{'} =\]

\[= 6x^{5} + 2 \bullet 2x = 6x^{5} + 4x\]

\[k = y^{'}(0) = 6 \bullet 0^{5} + 4 \bullet 0 = 0.\]

\[Уравнение\ касательной:\]

\[y(0) = 0^{6} = 0\]

\[y = 0 + 0(x - 0) = 0.\]

\[Ответ:\ \ y = 0.\]

\[2)\ y = x^{4}\text{\ \ }и\ \ y = x^{3} - 3x^{2}\]

\[x^{4} = x^{3} - 3x^{2}\]

\[x^{4} - x^{3} + 3x^{2} = 0\]

\[x^{2} \bullet \left( x^{2} - x + 3 \right) = 0\]

\[x = 0\]

\[x^{2} - x + 3 = 0\]

\[D = 1 - 12 = - 11 < 0\]

\[Первая\ касательная:\]

\[1)\ y^{'}(x) = \left( x^{4} \right)^{'} = 4x^{3}\]

\[k = y^{'}(0) = 4 \bullet 0^{3} = 0.\]

\[2)\ y^{'}(x) = \left( x^{3} \right)^{'} - 3 \bullet \left( x^{2} \right)^{'} =\]

\[= 3x^{2} - 3 \bullet 2x = 3x^{2} - 6x\]

\[k = y^{'}(0) = 3 \bullet 0^{2} - 6 \bullet 0 = 0.\]

\[Уравнение\ касательной:\]

\[y(0) = 0^{4} = 0\]

\[y = 0 + 0(x - 0) = 0.\]

\[Ответ:\ \ y = 0.\]

\[3)\ y = (x + 2)^{2}\text{\ \ }и\ \ y = 2 - x^{2}\]

\[(x + 2)^{2} = 2 - x^{2}\]

\[x^{2} + 4x + 4 - 2 + x^{2} = 0\]

\[2x^{2} + 4x + 2 = 0\]

\[2\left( x^{2} + 2x + 1 \right) = 0\]

\[(x + 1)^{2} = 0\]

\[x + 1 = 0\]

\[x = - 1.\]

\[1)\ y^{'}(x) = {(x + 2)^{2}}^{'} = 2(x + 2)\]

\[k = y^{'}( - 1) = 2( - 1 + 2) =\]

\[= 2 \bullet 1 = 2.\]

\[2)\ y^{'}(x) = (2)^{'} - \left( x^{2} \right)^{'} =\]

\[= 0 - 2x = - 2x\]

\[k = y^{'}( - 1) = - 2 \bullet ( - 1) = 2.\]

\[Уравнение\ касательной:\]

\[y( - 1) = ( - 1 + 2)^{2} = 1^{2} = 1\]

\[y = 1 + 2(x + 1) =\]

\[= 1 + 2x + 2 = 2x + 3.\]

\[Ответ:\ \ y = 2x + 3.\]

\[4)\ y = x(2 + x)\text{\ \ }и\ \ y = x(2 - x)\]

\[x(2 + x) = x(2 - x)\]

\[2x + x^{2} = 2x - x^{2}\]

\[2x^{2} = 0\]

\[x = 0.\]

\[1)\ y^{'}(x) = \left( x(2 + x) \right)^{'} =\]

\[= (2x)^{'} + \left( x^{2} \right)^{'} = 2 + 2x\]

\[k = y^{'}(0) = 2 + 2 \bullet 0 = 2.\]

\[2)\ y^{'}(x) = \left( x(2 - x) \right)^{'} =\]

\[= (2x)^{'} - \left( x^{2} \right)^{'} = 2 - 2x\]

\[k = y^{'}(0) = 2 - 2 \bullet 0 = 2.\]

\[Уравнение\ касательной:\]

\[y(0) = 0 \bullet (2 + 0) = 0\]

\[y = 0 + 2(x - 0) = 2x.\]

\[Ответ:\ \ y = 2x.\]